Problem: Solve for $x$ and $y$ using substitution. ${-6x+4y = -6}$ ${x = 3y+8}$
Explanation: Since $x$ has already been solved for, substitute $3y+8$ for $x$ in the first equation. ${-6}{(3y+8)}{+ 4y = -6}$ Simplify and solve for $y$ $-18y-48 + 4y = -6$ $-14y-48 = -6$ $-14y-48{+48} = -6{+48}$ $-14y = 42$ $\dfrac{-14y}{{-14}} = \dfrac{42}{{-14}}$ ${y = -3}$ Now that you know ${y = -3}$ , plug it back into $\thinspace {x = 3y+8}\thinspace$ to find $x$ ${x = 3}{(-3)}{ + 8}$ $x = -9 + 8$ ${x = -1}$ You can also plug ${y = -3}$ into $\thinspace {-6x+4y = -6}\thinspace$ and get the same answer for $x$ : ${-6x + 4}{(-3)}{= -6}$ ${x = -1}$